# Showing that $\lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\sum_{p<Q}1\right|=0$

As the title of the question susggests, I would like to show that

The "trivial bound is that"

\begin{align*} \lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n

where the equality is obtained by noting that $$\frac{1}{p}$$ numbers are multiplies of $$p$$, so the expected value of $$\sum_{\substack{p is exactly $$\sum_{p. Thus, we are looking only for a "$$o(\cdot)$$" improvement. The first thought of mine would be to note that since $$\sum_{\substack{p is an additive function and so by the Turan-Kubilius inequality

$$\sum_{n

The issue is, this inequality is worse than the trivial one since apply Cauchy-Shwartz we get that this inequality yields

\begin{align*} \frac{1}{N\pi(Q)}\sum_{n

where by the PNT this last term is on the order of $$\sqrt{\frac{Q^2}{\log(Q)}}=\frac{Q}{\sqrt{\log(Q)}}$$. This inequality would indicate to us that the sum would $$\mathit{diverge}$$, namely

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n

I have proved that the sum can't go to zero too fast, and specifically that

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n

for any $$\epsilon>0$$. It does, however, feel natural that this sum should be at least $$o(1)$$.

• I think the third equation should read: $\cdots \leq \frac{1}{\sqrt{N}\pi(Q)} \sqrt{\frac{1}{N}\cdots}$ which implies that the limit as $N$ tends to infinity is $0$. – David Tweedle Oct 28 '20 at 2:21
• @DavidTweedle Are you sure that you accounted for the extra factor of $N$ that appears in the square root when you apply Cauchy-Shwartz? – Milo Moses Oct 28 '20 at 2:28
• The inner limit, $\lim_{N\to\infty}\frac 1{N\pi(Q)}\sum_{n<N}\left|\sum_{p<Q,\,p|n} p-\pi(Q)\right|$ is exactly $\frac1{M\pi(Q)}\sum_{n<M}\left|\sum_{p<Q,\,p|n} p-\pi(Q)\right|$, where $M$ is the product of all primes less than $Q$ since the summand in absolute values is $M$-periodic in $n$. – Anthony Quas Oct 28 '20 at 7:37
• If $n$ is uniformly distributed on the set $\{1,\ldots,M\}$, then the events $p|n$ are mutually independent for all $p<Q$. This means you can rewrite the inner limit as $\frac 1{\pi(Q)}\mathbb E\left|\sum_{p<Q} X_p-\pi(Q)\right|$, where the $X_p$'s are independent random variables taking the value $p$ with probability $\frac 1p$ and 0 otherwise. – Anthony Quas Oct 28 '20 at 7:49
• @MiloMoses Yes, I made a mistake. Sorry about that! – David Tweedle Nov 10 '20 at 13:19

It seems that the conjecture is false, if I did not miss some asymptotc issue The essence of what follows is that I show that the interior limit exceeds any function of $$Q$$ tending to $$0$$.

For any $$t\geq 1$$, denote by $$p_t(Q)$$ the density of those $$n$$ divisible by at least one $$p$$ with $$t\pi(Q). We have \begin{align*} p_t(Q)&=1-\prod_{t\pi(Q) where the estimates work in the regime $$t\leq o(Q/\pi(Q))=o(\log Q)$$. (More precisely, in any such regime this equivalence is uniform over $$t\leq o(\log Q)$$, as $$q\to\infty$$.)

Assume now that your conjecture is true, i.e., that $$f(Q):=\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n Take $$N$$ to be a multiple of $$Q!$$. Notice that at least $$p_t(Q)N$$ of the summands are larger than $$(t-1)\pi(Q)$$. Put $$q=\log Q$$. Summing up over $$t=2,3,\dots,qf(Q)=o(q)$$, we obtain \begin{align*} \frac1{N\pi(Q)}\sum_{n Since $$f(Q)=o(1)$$, the limit of the above expression as $$N\to\infty$$ cannot equal $$f(Q)$$.

Remark. The inequality $$(*)$$ holds, because if a number $$n$$ is accounted for $$x$$ times in the right sum, then its large prime divisor $$p$$ is larger than $$(x+1)\pi(Q)$$, so that $$\left|\sum_{\substack{px\pi(Q).$$

• Are you sure that there are no issues with counting numbers multiple times in the sum $\sum_{t=2}^{qf(Q)}p_t(Q)$? If it is the density of primes where at least on prime divides in that in those intervals, numbers which appear in the tightest interval will also appear in every other term. – Milo Moses Oct 28 '20 at 15:50
• It seems that I am; notice that there is no $\sim$ sign there. I've added a clarification in the Remark at the end. If you are unsure about some other steps --- feel free to ask, this is really sketchy; sorry for that. – Ilya Bogdanov Oct 28 '20 at 16:24
• Could you expand the last step in the first line of math where you claim that $1-\frac{\log(Q)}{\log(t\pi(Q))}$ is asymptotic to a much simpler expression (and at a uniform rate) – Milo Moses Oct 28 '20 at 16:37
• I've inserted a line into the equation. Is it really the place you got troubls, or perhaps you meany a different one? – Ilya Bogdanov Oct 29 '20 at 6:03
• Though it pains me to say it, I can't find any more questionable steps in your proof. Thank you very much for your help. – Milo Moses Oct 29 '20 at 15:13